Question

If $z=\sqrt{-7-24i}$, then $z$ is equal to

• A. $3-4i$
• B. $-3-4i$
• C. $-3+4i$
• D. $3+4i$

Answer 1

Answer: (A), (C)

$-3+4i$

Given : $z=\sqrt{-7-24i}$
Let $z=a+ib$, so
$z^2=-7-24i\Rightarrow (a+ib{)}^2=-7-24i\Rightarrow a^2-b^2+2i(ab)=-7-24i\Rightarrow a^2-b^2=-7,ab=-12$

Now,
$(a^2+b^2{)}^2=(a^2-b^2{)}^2+4a^2{b}^2\Rightarrow (a^2+b^2{)}^2=49+4(12{)}^2\Rightarrow (a^2+b^2{)}^2=625\Rightarrow a^2+b^2=25(\because a,b\in R)$

Using $a^2-b^2=-7,a^2+b^2=25$, we get
$a^2=9,b^2=16\Rightarrow a=\pm 3,b=\pm 4$
Using $ab=-12$, we get
$a=3,b=-4a=-3,b=4$
Hence, $z=3-4i,-3+4i$

Alternate solution :
We know that,
$\sqrt{a-ib}=\pm \left[\sqrt{\frac{1}{2}(\sqrt{a^2+b^2}+a}\right)-i\sqrt{\frac{1}{2}(\sqrt{a^2+b^2}-a}\left)\right]$
If $(b>0)$
Hence $z=\sqrt{-7-24i}=\sqrt{-7-\left(24\right)i}$
$z=\pm \left[\sqrt{\frac{1}{2}(\sqrt{7^2+{24}^2}-7}\right)-i\sqrt{\frac{1}{2}(\sqrt{7^2+{24}^2}+7}\left)\right]$
$z=\pm (3-4i)$
$\therefore z=-3+4i$ and $z=3-4i$